finding slope of the tangent by differentiating x and y with respect to theta
\(\frac{dx}{d\theta}=-3\sin\theta+3\cos^2\theta\sin\theta\)
\(\frac{dy}{d\theta}=3\cos\theta-3\sin^2\theta\cos\theta\)
Now dividing \(\frac{dy}{d\theta}\) and \(\frac{dx}{d\theta}\) to obtain the slope of tangent
\(\frac{dy}{dx}\)\(=\frac{3\cos\theta-3sin^2\theta\cos\theta}{-3\sin\theta+3cos^2\theta\sin\theta}\)\(=-\tan^3\theta\)
m(tangent) at theta is\(=-\tan^3\theta\)
normal is perpendicular to tangent so, m1m2 = – 1
m(normal) at theta is cot3θ
equation of tangent is given by y – y1 = m(tangent)(x – x1)
y - 3 sin θ + sin3 θ = -tan3 θ(x - 3cos θ + 3cos3 θ)
equation of normal is given by y – y1 = m(normal)(x – x1)
y - 3 sin θ + sin3 θ = cot3 θ(x - 3cos θ + 3cos3 θ)